In $\triangle A B C$, let $D, E$, and $F$ be the midpoints of the sides of the triangle, and let $P, Q,$ and $R$ be the midpoints of the corresponding medians, $AD ,B E,$ and $C F$, respectively, as shown in the figure at the right. Prove that the value of \[\frac{AQ^2 + A R^2 + B P^2 + B R^2 + C P^2+ C Q^2 }{A B^2 + B C^2 + C A^2}\] does not depend on the shape of $\triangle A B C$ and find that value. [asy] defaultpen(linewidth(0.7)+fontsize(10));size(200); pair A=origin, B=(14,0), C=(9,12), D=midpoint(C--B), E=midpoint(C--A), F=midpoint(A--B), R=midpoint(C--F), P=midpoint(D--A), Q=midpoint(E--B); draw(A--B--C--A, linewidth(1)); draw(A--D^^B--E^^C--F); draw(B--R--A--Q--C--P--cycle, dashed); pair point=centroid(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$P$", P, dir(40)*dir(point--P)); label("$Q$", Q, dir(40)*dir(point--Q)); label("$R$", R, dir(40)*dir(point--R)); dot(P^^Q^^R);[/asy]

Given an acute triangle $ABC$, let $l_a$ be the line passing $A$ and perpendicular to $AB$, $l_b$ be the line passing $B$ and perpendicular to $BC$, and $l_c$ be the line passing $C$ and perpendicular to $CA$. Let $D$ be the intersection of $l_b$ and $l_c$, $E$ be the intersection of $l_c$ and $l_a$, and $F$ be the intersection of $l_a$ and $l_b$. Prove that the area of the triangle $DEF$ is at least three times of the area of $ABC$.

Equiangular hexagon $ ABCDEF$ has side lengths $ AB \equal{} CD \equal{} EF \equal{} 1$ and $ BC \equal{} DE \equal{} FA \equal{} r$. The area of $ \triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $ r$? $ \textbf{(A)}\ \frac {4\sqrt {3}}{3} \qquad \textbf{(B)}\ \frac {10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac {17}{4} \qquad \textbf{(E)}\ 6$

In triangle $ABC$, angles $A$ and $B$ measure 60 degrees and 45 degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$, and $AT=24.$ The area of triangle $ABC$ can be written in the form $a+b\sqrt{c},$ where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c.$

A triangle has sides of length $\sqrt{13}$, $\sqrt{17}$, and $2 \sqrt{5}$. Compute the area of the triangle.

A convex polygon $\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.

The diagram below shows the regular hexagon $BCEGHJ$ surrounded by the rectangle $ADFI$. Let $\theta$ be the measure of the acute angle between the side $\overline{EG}$ of the hexagon and the diagonal of the rectangle $\overline{AF}$. There are relatively prime positive integers $m$ and $n$ so that $\sin^2\theta  = \tfrac{m}{n}$. Find $m + n$. [asy] import graph; size(3.2cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((-1,3)--(-1,2)--(-0.13,1.5)--(0.73,2)--(0.73,3)--(-0.13,3.5)--cycle); draw((-1,3)--(-1,2)); draw((-1,2)--(-0.13,1.5)); draw((-0.13,1.5)--(0.73,2)); draw((0.73,2)--(0.73,3)); draw((0.73,3)--(-0.13,3.5)); draw((-0.13,3.5)--(-1,3)); draw((-1,3.5)--(0.73,3.5)); draw((0.73,3.5)--(0.73,1.5)); draw((-1,1.5)--(0.73,1.5)); draw((-1,3.5)--(-1,1.5)); label("$ A $",(-1.4,3.9),SE*labelscalefactor); label("$ B $",(-1.4,3.28),SE*labelscalefactor); label("$ C $",(-1.4,2.29),SE*labelscalefactor); label("$ D $",(-1.4,1.45),SE*labelscalefactor); label("$ E $",(-0.3,1.4),SE*labelscalefactor); label("$ F $",(0.8,1.45),SE*labelscalefactor); label("$ G $",(0.8,2.24),SE*labelscalefactor); label("$ H $",(0.8,3.26),SE*labelscalefactor); label("$ I $",(0.8,3.9),SE*labelscalefactor); label("$ J $",(-0.25,3.9),SE*labelscalefactor); [/asy]

In quadrilateral $ABCD$, $BC=8$, $CD=12$, $AD=10$, and $m\angle A= m\angle B = 60^\circ$. Given that $AB=p + \sqrt{q}$, where $p$ and $q$ are positive integers, find $p+q$.

Let $\triangle ABC$ be a triangle. Let $M$ be the midpoint of $BC$ and let $D$ be a point on the interior of side $AB$. The intersection of $AM$ and $CD$ is called $E$. Suppose that $|AD|=|DE|$. Prove that $|AB|=|CE|$.

Convex quadrilateral $ ABCD$ is inscribed in a circle, $ \angle{A}\equal{}60^o$, $ BC\equal{}CD\equal{}1$, rays $ AB$ and $ DC$ intersect at point $ E$, rays $ BC$ and $ AD$ intersect each other at point $ F$. It is given that the perimeters of triangle $ BCE$ and triangle $ CDF$ are both integers. Find the perimeter of quadrilateral $ ABCD$.

In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, $BN\perp AN$ and $\theta$ is the measure of $\angle BAC$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals [asy] size(230); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=14*dir(36), C=intersectionpoint(B--(9001,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b); draw(N--B--A--N--M--C--A^^B--M); markscalefactor=0.1; draw(rightanglemark(B,N,A)); pair point=N; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, S); label("$N$", N, dir(30)); label("$19$", (A+C)/2, dir(A--C)*dir(90)); label("$14$", (A+B)/2, dir(A--B)*dir(270)); [/asy] $\displaystyle \text{(A)} \ 2 \qquad \text{(B)} \ \frac{5}{2} \qquad \text{(C)} \ \frac{5}{2} - \sin \theta \qquad \text{(D)} \ \frac{5}{2} - \frac{1}{2} \sin \theta \qquad \text{(E)} \ \frac{5}{2} - \frac{1}{2} \sin \left(\frac{1}{2} \theta\right)$

Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP>CP$. Let $O_1$ and $O_2$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB=12$ and $\angle O_1 P O_2 = 120^\circ$, then $AP=\sqrt{a}+\sqrt{b}$ where $a$ and $b$ are positive integers. Find $a+b$.

Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with diameter $XY$. Segments $AY$ and $BX$ meet at $P$. Point $Z$ is the foot of the perpendicular from $P$ to line $XY$. Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$. Let $Q$ be the intersection of segments $AY$ and $XC$. Prove that \[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.\]

Let $ABC$ be a triangle with $AB = 4024$, $AC = 4024$, and $BC=2012$. The reflection of line $AC$ over line $AB$ meets the circumcircle of $\triangle{ABC}$ at a point $D\ne A$. Find the length of segment $CD$. [i]Ray Li.[/i]